What is the extraneous solution to these equations? $\dfrac{x^2 + 3}{x + 2} = \dfrac{-5x - 3}{x + 2}$
Multiply both sides by $x + 2$ $ \dfrac{x^2 + 3}{x + 2} (x + 2) = \dfrac{-5x - 3}{x + 2} (x + 2)$ $ x^2 + 3 = -5x - 3$ Subtract $-5x - 3$ from both sides: $ x^2 + 3 - (-5x - 3) = -5x - 3 - (-5x - 3)$ $ x^2 + 3 + 5x + 3 = 0$ $ x^2 + 6 + 5x = 0$ Factor the expression: $ (x + 2)(x + 3) = 0$ Therefore $x = -2$ or $x = -3$ At $x = -2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -2$, it is an extraneous solution.